Diffraction With Infrared Light

What happens when a wave passes through some opening? It diffracts – or you like, you could say “bends”. If the size of the opening is comparable to the wavelength of the wave, the bending can be noticeable. If you pass a wave (like light) through a whole bunch of openings, you can a nice and clear diffraction pattern. Why? First, there is something you need to know about waves. If you have two waves meeting together at the same place and the same time, they add together. You can try this yourself. Take some type of rope and stretch it out on the ground. If you move one end up and down really quickly, you can get a pulse wave traveling down the rope. Now here is the cool part. What if two people are on opposite ends of the rope and send down a pulse at the same time? What happens when these pulses meet? They add together like this:

When waves add together to make a larger amplitude, it is called constructive interference. If you shake the string to the side, you can try one pulse and the other an “inverse pulse”. When these meet, they will cancel like this:

This would be called destructive interference. Ok, back to the whole bunch of openings. A glass plate with really small lines would act just like a whole bunch of openings. We call this a diffraction grating and it makes a nice interference pattern. Let me see if I can make a sketch of waves passing through just two openings.

Yes. This is just a sketch. But here you can see that if there are two expanding waves, there are lines along which the constructively interfere. I didn’t draw it, but there are also lines that could show where the waves destructively interfere. If there were a screen on the other side of the slits, you would see a diffraction pattern. Here is an example with a red laser:

It turns out that there is a relationship between the spacing of these bright spots, the wavelength of the light and the slit spacing. When do you get a spot with constructive interference? It has to do with path length. If the light from one slit travels one whole wavelength more than the light from the other slit, then they will be back in phase and constructively interfere (make a bright spot). Time for another drawing. In this case, I will not show the waves but just the path the light takes going through two slits.

Using some simple geometry, I can find the path length difference between path 1 and path 2. And yes, this assumes that the point where the waves interfere is quite far away (so it is safe to assume these two paths are parallel). So, there will be constructive interference if:

Where λ is the wavelength of the light. Also, this is the angular location of the first bright spot to the side (there will also be a bright spot at θ = 0). The next bright spot will be where the path length is twice the wavelength. In general, I can write:

Really, the cool part is that this bright spot location depends on the wavelength. If you send several wavelengths through at the same time, they will make bright spots at different locations so that you can see what wavelengths are there.

On to the experiment

I now have 3 different wavelengths of lasers – blue, green and red. Here is a picture looking through a diffraction grating at a dot made by each of these lasers. For this picture, the diffraction grating is mounted on the camera.

All of these lasers list a wavelength.

Red: 640-680 nm

Green: 532 nm

Blue: 405 +/- 10 nm

That is how they are listed on the laser. Now for the plotting. I don’t know the angular position of these bright dots, but here is a diagram of what that would look like:

L is the distance from the diffraction grating to the screen (something quite large, like 1 or 2 meters – but a constant). y is the distance from the center bright spot to the next spot (much smaller, like a couple of centimeters). It isn’t a terrible approximation to say that the hypotenuse of this triangle is also about the length L. This means:

I don’t know the value for L, but I can proceed anyway. Actually, I don’t know the spacing between gratings (d) either. So, let me re-write the bright spot equation in terms of y:

For a given laser, &lambda, L and d are all constant. So, if I plot y vs. m, it should be a straight line. Here is that plot for the three lasers above:

And since I know the wavelengths, I can solve for L/d. All of the slopes give almost the same ratio of 0.43 – so that is good. Now, for something cool.

Infrared Diffraction

Ok, you have to admit, using the three different colored lasers is cool. Cool, but not that new. So what about using diffraction with infrared light? Yes, that would be cool, but we can’t really see infrared (well, there is a body hack for this). Even though we can’t see infrared, most video cameras can. Now wait. Don’t get too excited. Most video cameras can see near infrared. This is the infrared wavelengths used in things such as IR remotes. This is not the same wavelength of infrared that people associate with thermal radiation. Actually (before I forget), I like to remind you that using a camera is a great way to check and see if your IR remote is even working. Point it at your phone camera or something and press a button. In the camera, you should see a flickering light. I tried this with my phone and it worked with the front-facing camera, but not the other one. Some cameras include an IR-filter to block out these wavelengths. Back to the experiment. Here I am using the exact same diffraction grating with the same camera. The only difference is that I pointed the IR remote directly at the camera (for the laser pointers, I made a dot on a screen). Here is the image you would get (along with the other colors for comparison)

Those are two different IR remotes, but it looks like they are the same wavelength (since their bright spot positions are about the same). Here is the same plot above with the IR data added.

And assuming L and d are the same, this means the wavelength of this IR remote would be about 920 nm (920 x 10-9 m). For most people, the longest wavelength of red light is around 700 nm, so this would be in the IR range and in the “near IR”.